TITLE: Is there a "simplification" functor in algebraic topology?
QUESTION [38 upvotes]: Recall that a space (=CW complex) is called simple if it is connected, the fundamental group is abelian, and the fundamental group acts trivially on all higher homotopy groups.  Call Simp(X) a simplification of X if it is universal for maps from X to a simple space.  Does Simp(X) exist for any connected space X? This would give a higher analogue of abelianization of groups.
(A natural guess would be the loop suspension of X, but I'm pretty sure that doesn't work as the following example shows.  Let X be $\mathbb{R}\mathrm{P}^2$, and note that the $2$-type of $X$ can be described completely by saying that $\pi_1(X) = \mathbb{Z}/2$, $\pi_1(X) = \mathbb{Z}$, the action is the nontrivial one sending $n \mapsto -n$, and the gluing is by the nontrivial element of $H^3(\mathbb{Z}/2,\mathbb{Z})$.  There is a simple space $Y$ which is a $2$-type where $\pi_1(X) = \mathbb{Z}/2$, $\pi_2(X) = \mathbb{Z}/2$, the action is trivial, and the gluing is by the nontrivial element of $H^3(\mathbb{Z}/2,\mathbb{Z}/2)$.  There's a natural map from X to Y since the nontrivial action on $\mathbb{Z}$ becomes trivial when you kill $2$.  On the other hand, $\pi_2(\mathbb{R}\mathrm{P}^2) = \mathbb{Z}$ maps to $\pi_2(\Omega \Sigma \mathbb{R}\mathrm{P}^2) = \mathbb{Z}/4$ sending $1 \mapsto 2$, and so the surjective map $\mathbb{Z} = \pi_2(X) \rightarrow \pi_2(Y) = \mathbb{Z}/2$ cannot factor through $\pi_2(\Omega \Sigma \mathbb{R}\mathrm{P}^2)$.)

REPLY [3 votes]: Sorry to resurrect an old question, but I just wanted to expand here on an observation that Qiaochu made above (in the form of a question, but I think he was just playing Jeopardy!).
 Let $X$ be a connected space, and suppose that it has a simplification $X \to X^s$. Since Eilenberg-MacLane spaces are simple, the map $X \to X^s$ is a (co)homology equivalence in all degrees. Since every local coefficient system on $X^s$ is trivial, this implies that $X \to X^s$ is an acyclic map. It also must kill the perfect radical of $\pi_1(X)$. But the plus construction $X \to X^+$ is the unique acyclic map killing the perfect radical of $\pi_1(X)$. Therefore $X^s  = X^+$.
If $X^+$ happens to be simple, then $X \to X^+$ is a simplification, simply because any map $X \to Y$ with $Y$ simple must kill the perfect radical (which in this case must coincide with the commutator) of $\pi_1(X)$, and $X \to X^+$ is the universal map which does this.
So among the spaces $X$ that do have a simplification are:

$X$ such that $\pi_1(X)$ is perfect
$X$ such that $X^+$ is an $H$-space (e.g. all the cases used in algebraic $K$-theory)

We can also conclude that if $X$ has a simplification $X \to X^s$, then it factors through the plus construction $X \to X^+ \to X^s$, and $X^s$ is also the simplification of $X^+$. So in asking which spaces $X$ admit a simplification, we can reduce to the case where $\pi_1(X)$ is solvable (at least under suitable finiteness assumptions).
Warning: From here on, this discussion grows increasingly aimless.
Because Eilenberg-MacLane spaces are simple, a simplification is a (co)homology isomorphism.
Assume $X^s$ exists, and write $G = \pi_1(X), G^s = \pi_1(X^s)$. If $G \to G^s$ is not surjective, then it lifts to some cover $\overline{X^s}$ of $X^s$. On homology, $X^s$ is a retract of $\overline{X^s}$, but both spaces have abelian fundamental group so it's also a retract on $\pi_1$. But the map on $\pi_1$ is injective, so it is an isomorphism. Hence $G \to G^s$ is surjective after all, and is precisely the quotient by the commutator subgroup, i.e. we have $G^s = G^{ab}$.
So the cofiber of $X \to X^s$ is acyclic and is also simply-connected, so is contractible. Since the 1-truncation functor preserves cofiber sequences, the map $BG \to BG^{ab}$ likewise has trivial cofiber. So $BG \to BG^{ab}$ is a homology isomorphism.
I'm not sure where to go from there, so instead, let's pick off a special case. Assume that $G$ is abelian, so $G = G^{ab}= G^s$. Then by comparing fiber sequences $\tilde X \to X \to BG$ and $\tilde X^s \to X^s \to BG$ (where tilde's denote universal covers), we see that $H_\ast(\tilde X^s) = H_\ast(\tilde X)_G$ and in addition, the map $H_\ast(\tilde X) \to H_\ast(\tilde X)_G$ is an equivalence on $G$-homology, for any constant coefficients. So for example, if $G$ is a finitely-generated torsion-free abelian group and $H_\ast(X)$ is finitely-generated in each degree over $G$, then I think we can conclude that $X$ was already at least weakly simple (i.e. has trivial $\pi_1$-action on $H_\ast$).